Read the passage carefully and answer the Questions.

$KMnO_4$ is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$ to give a dark- green manganate ion which disproportionate to give permanganate as follows.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.

Oxidation states of Mn in $MnO_2, KMnO_4$ and $K_2MnO_4$ are

+4, +7, and +6 respectively

The correct answer is Option (4) → +4, +7, and +6 respectively

• MnOâ‚‚: Let the oxidation state of Mn = x
• Oxygen = –2  so 2 O = –4

$x+2(−2)=0  ⟹  x−4=0  ⟹  x=+4$

• KMnOâ‚„: Let the oxidation state of Mn = y
• K = +1  and Oâ‚„ = –8

$+1+y+4(−2)=0  ⟹  1+y−8=0  ⟹  y=+7$

• Kâ‚‚MnOâ‚„: Let the oxidation state of Mn = z
• 2K = +2 and Oâ‚„ = –8

$2(+1)+z+4(−2)=0  ⟹  2+z−8=0  ⟹  z=+6$

So, the oxidation states are MnOâ‚‚ = +4, KMnOâ‚„ = +7, Kâ‚‚MnOâ‚„ = +6.